1
$\begingroup$

I am currently reading the chapter "Fourier Optics" in the book "Fundamentals of Photonics" by Saleh and Teich. However I am not able to follow one specific mathematical derivation.

On page 111 the transfer function of free space is derived $$ H(\nu_x, \nu_y) = \text{exp}(-j 2 \pi d \sqrt{\lambda^{-2} - \nu_x^2 - \nu_y^2}).$$

$d$ is the distance the light travels from the input plane to the output plane. $\lambda$ is the wavelength and $\nu_x$ and $\nu_y$ are the spatial frequency components.

After that this formula is being simplified by using the fresnel approximation, for which it is assumed, that the frequency components $\nu_x$ and $\nu_y$ in the input wave are much smaller than the system bandwidth $\lambda^{-1}$. The resulting approximated transfer function is $$ H_{\text{Fresnel}}(\nu_x, \nu_y) = \text{exp}(j \pi \lambda d (\nu_x^2 + \nu_y^2)) \cdot \text{exp}(-j k d).$$

This still makes sense to me, everything is fine so far. However after that they derive the impulse response of the system by applying the inverse fourier transform to the transfer function $H_{\text{Fresnel}}$. The resulting function is $$h(x,y) \approx \dfrac{j}{\lambda d} \cdot \text{exp}(-j k d) \cdot \text{exp}(-j k \dfrac{x^2+y^2}{2 d}).$$

And honestly, I have absolutely no idea how they come to that expression. The inverse fourier is $$h(x, y) \approx \int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} H_{\text{Fresnel}}(\nu_x, \nu_y) \cdot \text{exp}(-j 2 \pi (\nu_x x + \nu_y y)) d\nu_x d\nu_y.$$

Small annotation: Out of some reason they flipped the signs in the fourier transform in contrast to the standard notation.

So the core question is: How did they solve this integral? There is a correspondence table at the end of the book, but I have no clue how this should help. Fourier transformation correspondence table Kind regards

$\endgroup$
3
  • $\begingroup$ see line 206 in table of en.wikipedia.org/wiki/Fourier_transform (i think you have a sign error in the equation for $H_{fresnel}$) $\endgroup$ – hyportnex Apr 4 '20 at 15:11
  • $\begingroup$ @hyportnex You were right about the sign error, but the suggested correspondence doesnt work because $Re(\alpha) = 0$ $\endgroup$ – cakelover Apr 5 '20 at 9:40
  • 1
    $\begingroup$ Have you heard/learned of analytic continuation? Line "206" as written is true for any complex $\alpha$ that has $Re[\alpha] \ge 0$ otherwise the integral is not convergent. It is standard practice in Fourier (or Laplace) transform analysis to develop a formula that is valid in the open upper (or right) half-plane whose boundary is the real (or imaginary) axis onto which analytic continuation gets you the answer you are seeking. $\endgroup$ – hyportnex Apr 5 '20 at 16:54
1
$\begingroup$

Remeber that for $a>0$ the Fresnel integral is $$ \int_{-\infty}^{\infty} e^{iax^2} = e^{i\pi/4} \sqrt{\frac \pi{a}}, $$
because of the need to push the contour off the real axis with $x= e^{i\pi/4}t$. Your integral has the product of two Fresnel integrals $x$ times $y$ and so you have $$ \left[e^{i\pi/4} \sqrt{\frac \pi{a}}\right]^2= \frac{\pi i}{a}. $$

$\endgroup$
0
$\begingroup$

I think I was able to solve the problem by applying the same method as mentioned here. However my solution still differs by a constant factor from the solution in the book, so maybe it is still not completely right.

If you look at $h(x, y)$ one can see easily that it can be separated as $$h(x, y) = K \cdot f(x) \cdot f(y).$$ with $f(x) = e^{\dfrac{-j k x^2}{2 d}} = e^{j a x^2}, a = \dfrac{-k}{2 d}$ and $K = e^{-j k d}$.

So if we know how to fourier transform $f(x)$, the problem is more or less solved.

If we differentiate $f(x)$, we get the following equation $$\dfrac{d f(x)}{dx} = f(x) \cdot 2 j a x.$$

Lets fourier transform it with the known correspondences $$j \omega F(\omega) = \dfrac{d F(\omega)}{d \omega} 2ja.$$

This gives us $$\dfrac{d F(\omega)}{d \omega} = F(\omega) \cdot \dfrac{\omega}{2 a j}.$$

We can see, that $$F(\omega) = \text{exp}(\dfrac{-j \omega^2}{4a})$$ is a solution of the differential equation.

No we can resubstitute $a$ and $k = \dfrac{2 \pi}{\lambda}$ $$F(\omega) = \text{exp}(\dfrac{j \omega^2 d \lambda}{4 \pi})$$ and with $\omega = 2\pi \nu$ we get $$F(\nu) = \text{exp}(j \pi \lambda d \nu^2).$$

Resubstituting everything gives us $$H(\nu_x, \nu_y) = \dfrac{j}{\lambda d} \text{exp}(-j k d) \text{exp}(j \lambda d \pi (\nu_x^2+\nu_y^2)).$$

Out of some reason the factor $\dfrac{j}{\lambda d}$ is still wrong, but that's the best answer i could derive.

Kind Regards

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.